String Matching and Suffix Tree

Gusfield Ch1-7 EECS 458 CWRU Fall 2004

BLAST (Altschul et al’90) • Idea: true match alignments are very likely to contain a short segment that identical (or very high score). • Consider every substring (seed) of length w, where w=12 for DNA and 4 for protein. • Find the exact occurrence of each substring (how?) • Extend the hit in both directions and stop at the maximum score.

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Problems • Pattern matching: find the exact occurrences of a given pattern in a given structure ( string matching) • Pattern recognition: recognizing approximate occurences of a given patttern in a given structure (image recognition) • Pattern discovery: identifying significant patterns in a given structure, when the patterns are unknown (promoter discovery)

Definitions • • • • • •

String S[1..m] Substring S[i..j] Prefix S[1..i] Suffix S[i..m] Proper substring, prefix, suffix Exact matching problem: given a string P called pattern, and a long string T called text, find all the occurrences of P in T.

Naïve method • Align the left end of P with the left end of T • Compare letters of P and T from left to right, until – Either a mismatch is found (not an occurrence – Or P is exhausted (an occurrence)

• Shift P one position to the right • Restart the comparison from the left end of P • Repeat the process till the right end of P shifts past the right end of T • Time complexity: worst case θ(mn), where m=|P| and n=|T| • Not good enough!

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Speedup • Ideas: – When mismatch occurs, shift P more than one letter, but never shift so far as to miss an occurrence – After shifting, ship over parts of P to reduce comparisons – Preprocessing of P or T

Fundamental preprocessing • Can be on pattern P or text T. • Given a string S (|S|=m) and a position i >1, define Zi: the length of longest common prefix of S and S[i..m] • Example, S=abxyabxz zi 1

zi

i

Fundamental preprocessing Zi

0 0 0 3 0 0 0

a b x y a b x z a b x y a b x z a b x y a b x z a b x a b x

y

y

a b x

a b x

z

z

a b x y a b x z a b x y a b x z a b x y a b x z

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Fundamental preprocessing • Intention: – Concatenate P and T, inserted by an extra letter $: S=P$T – Every i, Zi≤|P| – Every i>|P|+1 and Zi=|P| records the occurrences of P in T

• Question: running time to compute all the Zs? The naïve method according to the definition runs in θ((m+n)2) time!

Fundamental preprocessing • Goal: linear time to compute all the Zs • Z-box: for Zi>0, it is the box starting at i with length Zi.(ending at i+Zi-1), • ri: the rightmost end of a Zj-box (j+Zj-1) for all 11. • li: the left node of Zj-box ending at ri α

α

1

Zli

li

i

ri

Fundamental preprocessing •

Computing Zk: – Given Zi for all 10) 2. k≤r: k is in the Z-box starting at l (substring S[l..r]), therefor S[k]=S[k-l+1], S[k+1]=S[k-l+2], …, S[r]=S[Zl]. In other words, Zk≥min{Zk-l+1,r-k+1}

α 1

k-l+1

α

β Zl

l

β k

r

4

Fundamental preprocessing • A) Zk-l+1< (r-k+1): Zk=Zk-l+1, and r, l remain unchanged • B) Zk-l+1 ≥ (r-k+1): Zk≥ (r-k+1) and start comparison between S[r+1] and S[r-k+2] until a mismatch is found (updating r and l accordingly if Zk≥ r-k+1) α k-l+1

1

β 1

α k-l+1

α

β Zl

l

k α

β Zl

β

l

r β ?

k

r

Fundamental preprocessing •

Conclusions: 1. Zk is correctly computed 2. There are a constant number of operations besides comparisons for each k – |S| iterations – Whenever a mismatch occurs, the iteration terminates – Whenever a match occurs, r is increased

3. In total at most |S| mismatches and at most |S| matches 4. Running time θ(|S|) and space θ (|S|)

Fundamental preprocessing • Th: there is a θ(n+m)-time and space algorithm which finds all the occurrences of P in T, where m=|P| and n=|T|. • Notes: – Alphabet-independent – Space requirement can be reduced to θ(m) – Not well suited for multiple patterns searching – Strictly linear, every letter in T has to be compared at least once

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Projects • Topics • Meeting: 3 times, as a group • Presentations: 25 minutes/student (~20m talk + 5m questions) • Term paper: single space, 11pt, 1in margin. 5-6p, 9-10p 10-12p, exclude references

The Boyer-Moore algorithm: an example • P=abxabxab, T=daaabxababxabxab daaabx ababx abx ab abx abx ab abx abx ab daaabx ababx abx ab abx abx ab abx abx ab

The Boyer-Moore algorithm • Rule 1: right-to-left comparison • Rule 2: Bad character rule – For each x∈∑, R(x) denotes the right-most occurrence of x in P (0 if doesn’t appear) – When a mismatch occurs, T[k] against P[i], shift P right by max{1, i-R(T[k])} places. This takes T[k] against P[R(T[k])] – |∑| space to store R-values

• Rule 3: good suffix rule

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The Boyer-Moore algorithm T

z

P before shift

t’

x

t

y

t

z

P after shift

t’

t

y

y

t

The Boyer-Moore algorithm • Rule 3: good suffix rule – When a mismatch occurs, T[k] against P[i] – Find the rightmost occurrence of P[(i+1)..m] in P such that the letter to the left differs P[i] – Shift P right such that this occurrence of P[(i+1)..m] is against T[(k+1)..(m+k-i)] – If there is no occurrence of P[(i+1)..m], find the longest prefix of P matches a suffix of P[(i+1)..m], shift P right such that this prefix is against the corresponding suffix

Preprocessing for the good suffix rule • Let L(i) denote the largest position less than m such that string P[i..m] matches a suffix of P[1..L(i)] • Let N(j) denote the longest suffix of substring P[1..j] that is also a suffix of P • Recall Zi the length of longest substring of P starts I and matches a prefix of S. t

L(i)

t i

L(i)

zi

zi 1

zi

i N(j)

N(j) m

m-N(j)+1

j

1

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Preprocessing for the good suffix rule • Thm: L(i) is the largest index j less than m such that N(j)≥|P[i..m]|=m-i+1. L(i) 1

L(i)

i

m

N(j)

N(j) 1

m-N(j)+1

j

m

The Knuth-Morris-Pratt Algorithm • History: – Best known – Not the method of choice, inferior in practice – Can be generalized for multiple string matching

• Preprocessing P • Example: *

a b x y a b x z w a b x y a b x z w

KMP • Idea: – left to right comparison, – Shift P more places without missing occurrence

• A prefix of P matches a proper suffix of P[1..i] and the next letters do not match! • Define si of P, 2